| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Geometric properties using vectors |
| Difficulty | Moderate -0.3 This is a straightforward vectors question testing collinearity and perpendicularity. Part (a) requires finding a scalar parameter using the condition that vectors AB and AC are parallel (standard technique). Part (b) uses the perpendicular condition (dot product = 0). Both are routine applications of basic vector properties with minimal algebraic manipulation, making this slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{p-1}{1-2} = \frac{-3-1}{4-2}\) or \(\frac{p+3}{1-4} = -2\) | M1 | Correct equation involving ratios seen. Or \(y = -\frac{1}{2}x + \frac{5}{2}\) correct, and substitute \(y=1\). Or clear correct diagram drawn, from \(x=-3\) to \(x=3\). Allow M1A1 with unclear working or no working |
| \(p = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = \binom{4}{-2}\), \(\overrightarrow{BD} = \binom{q-1}{-1}\), \(\overrightarrow{DA} = \binom{-3-q}{3}\) | M1 | Attempt to find vectors along 2 or 3 sides. Allow errors |
| \((-3-q)^2 + 3^2 = (q-1)^2 + 1 + 16 + 4\) | M1 | Their \(DA^2 = AB^2 + BD^2\) or \(\binom{4}{-2}\cdot\binom{q-1}{-1} = 0\) ft their \(\overrightarrow{AB}\) & \(\overrightarrow{BD}\) |
| \(q = 0.5\) | A1 | Must follow from correct working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Gradient of \(AB = -\frac{1}{2}\), gradient of \(BD = 2\) | ||
| \(BD\) is \(y-2 = 2(x-1)\) or \(y = 2x+c\) & \(c=0\) | M1 | Attempt find gradient and equation of \(BD\). Allow errors |
| \(BD\) is \(y = 2x\) | A1 | |
| When \(y=1\), \(x=0.5\); \(q = 0.5\) | A1 |
## Question 2:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{p-1}{1-2} = \frac{-3-1}{4-2}$ or $\frac{p+3}{1-4} = -2$ | M1 | Correct equation involving ratios seen. Or $y = -\frac{1}{2}x + \frac{5}{2}$ correct, and substitute $y=1$. Or **clear** correct diagram drawn, from $x=-3$ to $x=3$. Allow M1A1 with unclear working or no working |
| $p = 3$ | A1 | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = \binom{4}{-2}$, $\overrightarrow{BD} = \binom{q-1}{-1}$, $\overrightarrow{DA} = \binom{-3-q}{3}$ | M1 | Attempt to find vectors along 2 or 3 sides. Allow errors |
| $(-3-q)^2 + 3^2 = (q-1)^2 + 1 + 16 + 4$ | M1 | Their $DA^2 = AB^2 + BD^2$ or $\binom{4}{-2}\cdot\binom{q-1}{-1} = 0$ ft their $\overrightarrow{AB}$ & $\overrightarrow{BD}$ |
| $q = 0.5$ | A1 | Must follow from correct working seen |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of $AB = -\frac{1}{2}$, gradient of $BD = 2$ | | |
| $BD$ is $y-2 = 2(x-1)$ or $y = 2x+c$ & $c=0$ | M1 | Attempt find gradient and equation of $BD$. Allow errors |
| $BD$ is $y = 2x$ | A1 | |
| When $y=1$, $x=0.5$; $q = 0.5$ | A1 | |
---2 Points $A$ and $B$ have position vectors $\binom { - 3 } { 4 }$ and $\binom { 1 } { 2 }$ respectively.\\
Point $C$ has position vector $\binom { p } { 1 }$ and $A B C$ is a straight line.
\begin{enumerate}[label=(\alph*)]
\item Find $p$.
Point $D$ has position vector $\binom { q } { 1 }$ and angle $A B D = 90 ^ { \circ }$.
\item Determine the value of $q$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q2 [5]}}